6.042/18.062J Mathematics for Computer Science April 14, 2005 Srini Devadas and Eric Lehman Problem Set 9 Solutions Due: Monday, April 25 at 9 PM Problem 1. There are three coins: a penny, a nickel, and a quarter. When these coins are flipped: • The penny comes up heads with probability 1/3and tails with probability 2/3. • The nickel comes up heads with probability 3/4and tails with probability 1/4. • The quarter comes up heads with probability 3/5and tails with probability 2/5. Assume that the way one coin lands is unaffected by the way the other coins land. The goal of this problem is to determine the probability that an odd number of coins come up heads. For this first problem, we’ll closely follow the four­step procedure for solving probability problems described in lecture. Your solution should include a tree diagram. (a) What is the sample space for this experiment? Solution. We can regard each outcome as a triple indicating the orientation of the penny, nickel, and quarter. For example, the triple (H, T, H) is the outcome in which the penny is heads, the nickel is tails, and the quarter is heads. The sample space is the set of all such triples: {H, T } 3. (b) What subset of the sample space is the event that an odd number of coins come up heads? Solution. The event that an odd number of coins come up heads is the subset: {(H, H, H), (H, T, T ), (T, H, T ), (T,T,H)} (c) What is the probability of each outcome in the sample space? Solution. Edges in the tree diagram are labeled with the probabilities given in the problem statement. The probability of each outcome is the product of the probabil­ ities along the corresponding root­to­leaf path. The resulting outcome probabilities are noted in the tree diagram. 2 Problem Set 9 H 3/5 × 9/60 H ¨XX ¨ XX 3/4 ¨ 2/5 XX 6/60 ¨¨ T H¨ H H H HH 3/5 3/60 1/4 H T HX XXX 1/3 2/5 XX × 2/60 T @ H @ 2/3 3/5 18/60 @ H X @ ¨¨ XXX 3/4 ¨ 2/5 XX × 12/60 T @ ¨ T @¨¨ penny H HH H H 3/5 × 6/60 1/4 H T HX XXX 2/5 XX 4/60 nickel T quarter odd? prob. (d) What is the probability that an odd number of coins come up heads? Solution. The probability of an event is the sum of the probabilities of the outcomes in that event. In this case: Pr (odd number of heads) = Pr ({(H, H, H), (H, T, T ), (T, H, T ), (T,T,H)}) = Pr ((H, H, H)) + Pr ((H, T, T )) + Pr ((T, H, T )) + Pr ((T,T,H)) 9 2 12 6 = + + + 60 60 60 60 29 = 60 Problem 2. Professor Plum, Mr. Green, and Miss Scarlet are all plotting to shoot Colonel Mustard. If one of these three has both an opportunity and the revolver, then that person shoots Colonel Mustard. Otherwise, Colonel Mustard escapes. Exactly one of the three has an opportunity with the following probabilities: Pr (Plum has opportunity) = 1/6 Pr (Green has opportunity) = 2/6 Pr (Scarlet has opportunity) = 3/6 Exactly one has the revolver with the following probabilities, regardless of who has an opportuntity: Pr (Plum has revolver) = 4/8 Pr (Green has revolver) = 3/8 Pr (Scarlet has revolver) = 1/8 Problem Set 9 3 (a) Draw a tree diagram for this problem. Indicate edge and outcome probabilities. Solution. P 4/48 4/8 G H 3/48 HH 3/8 H 1/8H H 1/48 P S 1/6 P 8/48 4/8 G G 6/48 J HH 2/6 H 3/8 J HH J 1/8 H 2/48 J 3/6 S J J P S 4/8 12/48 J J G JJ 9/48 HH opportunity H 3/8 HH 1/8 H 3/48 S revolver prob. (b) What is the probability that Colonel Mustard is shot? Solution. Denote each outcome with a pair indicating who has the opportunity and who has the revolver. In this notation, the event that Colonel Mustard is shot consists of all outcomes where a single person has both: {(P,) P, (G, G)(,S, S)} n The probability of this event is the sum of the outcome probabilities: Pr{ (P, P ), (G, G)(,S, S)}) = Pr ((P,)) P + Pr ((G, G)) +)) Pr ((S, S = 4/48 + 6/48 + 3/48 = 13/48 (c) What is the probability that Colonel Mustard is shot, given that Miss Scarlet does not have the revolver? Solution. Let Sbe the event that Colonel Mustard is shot, and let Nbe the event 4 Problem Set 9 that Miss Scarlet does not have the revolver. The solution is: PrS (∩ N) PrS ( | N) = PrN () Pr ((P,) P, (G, G)) = Pr ((P,) P, (P, G), (G,) P, (G, G), (S, P ), (S, G)) 4 6 48+ 48 =4 3 8 6 12 9 48+ 48 + +48 + 48 + 48 48 5 = 21 (d) What is the probability that Mr. Green had an opportunity, given that Colonel Mustard was shot? Solution. Let Gbe the event that Mr. Green has an opportunity, and again let Sbe the event that Colonel Mustard is shot. Then the solution is: PrG ( ∩ S) PrG ( | S) = PrS () Pr ((G, G)) = Pr ((P,) P, (G, G)(,S, S)) 6 48 =4 6 3 48+ 48 + 48 6 = 13 Problem 3. There are three prisoners in a maximum­security prison for fictional villains: the Evil Wizard Voldemort, the Dark Lord Sauron, and Little Bunny Foo­Foo. The parole board has declared that it will release two of the three, chosen uniformly at random, but has not yet released their names. Naturally, Sauron figures that he will be released to his 2 home in Mordor, where the shadows lie, with probability 3 . A guard offers to tell Sauron the name of one of the other prisoners who will be re­ leased (either Voldemort or Foo­Foo). However, Sauron declines this offer. He reasons that if the guard says, for example, “Little Bunny Foo­Foo will be released”, then his own 1 probability of release will drop to 2 . This is because he will then know that either he or Voldemort will also be released, and these two events are equally likely. Using a tree diagram and the four­step method, either prove that the Dark Lord Sauron has reasoned correctly or prove that he is wrong. Assume that if the guard has a choice of naming either Voldemort or Foo­Foo (because both are to be released), then he names one of the two uniformly at random. Solution. Sauron has reasoned incorrectly. In order to understand his error, let’s be­ gin by working out the sample space, noting events of interest, and computing outcome probabilities: Problem Set 9 5 F 1/3 × × × 1 1/3 F, S F ¨ 1/6 × × ¨¨ ¨ 1/2 F, V ¨¨ H @ 1/3 HH 1/2 @ HH V, S@ V H 1/6 × @ 1/3 @ 1 @ 1/3 × released V guard says prob. guard says Foo-foo Sauron ”Foo-foo” released released Define the events S, F , and “F ” as follows: “F ” = Guard says Foo-Foo is released F = Foo-Foo is released S = Sauron is released The outcomes in each of these events are noted in the tree diagram. Sauron’s error is in failing to realize that the event F (Foo-foo will be released) is dif- ferent from the event “F ” (the guard says Foo-foo will be released). In particular, the probability that Sauron is released, given that Foo-foo is released, is indeed 1/2: Pr (S ∩ F ) Pr (S | F ) = Pr (F ) 1 3 = 1 1 1 3 + 6 + 6 1 = 2 But the probability that Sauron is released given that the guard merely says so is still 2/3: Pr (S ∩ “F ”) Pr (S | “F ”) = Pr (“F ”) 1 3 = 1 1 3 + 6 2 = 3 So Sauron’s probability of release is actually unchanged by the guard’s statement. Problem 4. You shuffle a deck of cards and deal your friend a 5-card hand. 6 Problem Set 9 (a) Suppose your friend says, “I have the ace of spades.” What is the probablity that she has another ace? Solution. The sample space for this experient is the set of all 5­card hands. All �52 � outcomes are equally likely, so the probability of each outcome is 1/5 . Let Sbe the event that your friend has the ace of spades, and let A be the event that your friend has another ace. Our objective is to compute: PrA ( ∩ S) PrA ( | S) = PrS () The number of hands containing the ace of spades is equal to the number of ways to select 4 of the remaining 51 cards. Therefore: �51 � 4 PrS () = � 52� 5 The number of hands containing the ace of spades and at least one more ace is: �3�� 48� � �� 3� 48� �� 3� 48 |A ∩ S | = + + 1 3 2 2 3 1 Here the first term counts the number of hands with one additional ace, since there �3 � �48 � are 1 ways to choose the extra ace and 3 ways to choose the other cards. Sim­ ilarly, the second term counts hands with two additional aces, and the last term counts hands with all three remaining aces. In probability terms, we have: �3 ��48 � � 3��48 � �3 ��48 � 1 3 +2 2 + 3 1 PrA ( ∩ S) = �52 � 5 Substituting these results into our original equation gives the solution: �3 ��48 � �3 ��48 � �3 ��48 � 1 3 +2 2 + 3 1 PrAS ( | ) = �51 � =0.2214.